Question

find the margin of error for the given values of c, s, and n.
c = 0.98, s = 4, n = 12
click the icon to view the t - distribution table.
the margin of error is (round to three decimal places as needed).

Explanation:

Step1: Find the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 12\), then \(df=12-1 = 11\).

Step2: Locate the \(t\) - value

For a confidence level \(c = 0.98\), the two - tail area \(\alpha=1 - c=1 - 0.98 = 0.02\). Looking at the \(t\) - distribution table with \(df = 11\) and two - tail area \(\alpha=0.02\), the \(t\) - value \(t_{\alpha/2}=2.718\).

Step3: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(t_{\alpha/2}=2.718\), \(s = 4\), and \(n = 12\). Then \(E=2.718\times\frac{4}{\sqrt{12}}\).
First, \(\sqrt{12}\approx3.464\). Then \(\frac{4}{\sqrt{12}}\approx\frac{4}{3.464}\approx1.155\).
So \(E=2.718\times1.155 = 3.140\).

Answer:

\(3.140\)